SOLUTION: how do i find the remaining zeros of the function h(x)=2x^4+3x^3+96x^2+147x-98; zero: -7i

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Question 891388: how do i find the remaining zeros of the function
h(x)=2x^4+3x^3+96x^2+147x-98; zero: -7i
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39621) (Show Source):
You can put this solution on YOUR website!
"How....?"

Try this:

imaginary zeros occur in conjugate pair, so this means 7i is also a zero. This means there is a product , which becomes a quadratic factor with real integer coefficients.

Divide h(x) by using POLYNOMIAL DIVISION, unless you know another method, and then the quotient will be simpler to further reduce or factorize. (Meaning, finding the other two roots).

Answer by MathTherapy(10555) (Show Source):
You can put this solution on YOUR website!
how do i find the remaining zeros of the function
h(x)=2x^4+3x^3+96x^2+147x-98; zero: -7i

With one root being - 7i, the factors of: 2x^4 + 3x^3 + 96x^2 + 147x - 98 are:
(x + 7i)(x - 7i)(x + 2)(2x - 1).
Thus, besides - 7i, other roots are: , , and .